#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

typedef long long LL;

using namespace std;

const int N = 2000010;

int A,B,C,m;
LL s[N],b[N],bp[N];  // s[N]原数组 b[N]差分数组 bp[N]差分数组备份

int d[8][4] = {
    {0, 0, 0, 1},
    {0, 0, 1, -1},
    {0, 1, 0, -1},
    {0, 1, 1, 1},
    {1, 0, 0, -1},
    {1, 0, 1, 1},
    {1, 1, 0, 1},
    {1, 1, 1, -1}
};  // 二进制顺序记忆 奇数个1为-1

int op[N / 2][7];

int get(int i, int j, int k)  // 返回一维坐标
{
    return (i * B + j) * C + k;
}

bool check(int mid)
{
    memcpy(b, bp, sizeof b);
    for (int i = 1; i <= mid; i++)
    {
        int x1 = op[i][0], x2 = op[i][1], y1 = op[i][2], y2 = op[i][3], z1 = op[i][4], z2 = op[i][5], h = op[i][6];
        b[get(x1,y1,z1)] -= h;
        b[get(x1,y1,z2 + 1)] += h;
        b[get(x1,y2 + 1,z1)] += h;
        b[get(x1,y2 + 1,z2 + 1)] -= h;
        b[get(x2 + 1,y1,z1)] += h;
        b[get(x2 + 1,y1,z2 + 1)] -= h;
        b[get(x2 + 1,y2 + 1,z1)] -= h;
        b[get(x2 + 1,y2 + 1,z2 + 1)] += h;
    }

    memset(s, 0, sizeof s);  // s数组归零 通过b求前缀和 更新被攻击mid次后的s（战舰生命值）
    for (int i = 1; i <= A; i++)
        for (int j = 1; j <= B; j++)
            for (int k = 1; k <= C; k++)
            {
                s[get(i,j,k)] = b[get(i,j,k)];
                for (int u = 1; u < 8; u++)  // u从1开始，执行余下7次操作
                {
                    int x = i - d[u][0], y = j - d[u][1], z = k - d[u][2], c = d[u][3];
                    s[get(i,j,k)] -= s[get(x,y,z)] * c;  // 3维前缀和
                }

                if (s[get(i,j,k)] < 0) return true;  // 已经有战舰炸了 返回true
            }
    return false;
}

int main()
{
    cin >> A >> B >> C >> m;

    for (int i = 1; i <= A; i++)
        for (int j = 1; j <= B; j++)
            for (int k = 1; k <= C; k++)
                scanf("%lld", &s[get(i,j,k)]);  // 读入初始战舰生命值

    for (int i = 1; i <= A; i++)
        for (int j = 1; j <= B; j++)
            for (int k = 1; k <= C; k++)
                for (int u = 0; u < 8; u++)
                {
                    int x = i - d[u][0], y = j - d[u][1], z = k - d[u][2], c = d[u][3];
                    bp[get(i,j,k)] += s[get(x,y,z)] * c;  // 3维差分 
                }

    for (int i = 1; i <= m; i++)
        for (int j = 0; j < 7; j++)
            scanf("%d",&op[i][j]);  // 读入操作

    int l = 1, r = m;
    while (l < r)  // 二分查找
    {
        int mid = l + r >> 1;
        if (check(mid)) r = mid;  // check 判断当第mid次攻击发生，是否已经有战舰爆炸
        else l = mid + 1;
    }

    printf("%d", l);
    return 0;
}